MHB Is the integral over the first octant of a three-variable function continuous?

[Chris L T521]

Here's this week's problem!

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Background information: If a function $g$ of three variables is continuous and nonnegative on an unbounded region $S$ in $\Bbb{R}^3$, then the improper integral of $g$ over $S$ is defined by
\[\iiint\limits_S g\,dV = \lim_{c\to\infty} \iiint\limits_{S\cap B_c} g\,dV,\]
where $B_c$ is a ball of radius $c$ centered at any point $\mathbf{a}$ in $S$, provided that the limit exists.

Problem: Compute $\displaystyle\iiint\limits_S \frac{1}{(x^2+y^2+z^2+a^2)^{3/2}}\,dV$ where $S$ is the first octant ($x,y,z\geq 0$) and $a$ is a nonzero constant.

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[Chris L T521]

No one answered this week's problem. You can find my solution below.

[sp]Take $B_c$ to be a ball of radius $c$ centered at the origin. Then $S\cap B_c$ is the section of the sphere in the first octant. Converting the integral over $S\cap B_c$ into spherical coordinates, we see that
\[\begin{aligned}\int_0^{\infty}\int_0^{\infty}\int_0^{\infty}\frac{1}{(x^2+y^2+z^2+a^2)^{3/2}}\,dx\,dy\,dz &= \lim_{c\to\infty}\int_0^{\pi/2}\int_0^{\pi/2}\int_0^c \frac{\rho^2\sin\phi}{(\rho^2+a^2)^{3/2}}\,d\rho\,d\phi,d\theta\\ &= \left(\int_0^{\pi/2}\,d\theta\right)\left(\int_0^{\pi/2}\sin\phi\,d\phi\right)\left(\lim_{c\to\infty}\int_0^c \frac{\rho^2}{(\rho^2+a^2)^{3/2}}\,d\rho\right)\\ &= \frac{\pi}{2}\lim_{c\to\infty}\int_0^c \frac{\rho^2}{(\rho^2+a^2)^{3/2}}\,d\rho\end{aligned}\]

To evaluate the last integral, make the trig substitution $\rho = a\tan\theta\implies d\rho = a\sec^2\theta\,d\theta$. Therefore,
\[\begin{aligned}\int\frac{\rho^2}{(\rho^2+a^2)^{3/2}}\,d\rho &= \int\frac{a^3\tan^2\theta\sec^2\theta}{a^3\sec^3\theta}\,d\theta \\ &= \int\frac{\tan^2\theta}{\sec\theta}\,d\theta\\ &= \int \sec\theta+\cos\theta\,d\theta \\ &= \ln|\sec\theta+\tan\theta|+\sin\theta+C\\ &= \ln\left|\sqrt{\rho^2+a^2}+\rho\right|-\ln|a|+\frac{\rho}{\sqrt{\rho^2+a^2}}+C\\ &= \ln\left|\sqrt{\rho^2+a^2}+\rho\right|+\frac{\rho}{\sqrt{\rho^2+a^2}}+C\end{aligned}\]
Thus,

\[\begin{aligned} \frac{\pi}{2}\lim_{c\to\infty}\int_0^c\frac{\rho^2}{(\rho^2+a^2)^{3/2}}\,d\rho &= \frac{\pi}{2}\lim_{c\to\infty}\left.\left[\ln\left|\sqrt{\rho^2+a^2}+\rho\right|+\frac{\rho}{\sqrt{\rho^2+a^2}}\right]\right|_0^c\\ &= \frac{\pi}{2}\left[\lim_{c\to\infty}\ln\left|\sqrt{c^2+a^2}+c\right|+\frac{c}{\sqrt{c^2+a^2}} - \ln|a|\right] \end{aligned}\]

However, $\displaystyle\lim_{c\to\infty}\ln\left|\sqrt{c^2+a^2}+c\right|$ diverges to $\infty$. Therefore, we conclude that $\displaystyle\iiint\limits_S\frac{1}{(x^2+y^2+z^2+a^2)^{3/2}}\,dV = \lim_{c\to \infty}\iiint\limits_{S\cap B_c}\frac{1}{(x^2+y^2+z^2+a^2)^{3/2}}\,dV$ diverges to $\infty$.$\hspace{.25in}\clubsuit$[/sp]